Unit 4 The Bernoulli and Binomial Distributions


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1 PubHlth Bernoulli and Binomial Page 1 of 19 Unit 4 The Bernoulli and Binomial Distributions Topic 1. Review What is a Discrete Probability Distribution Statistical Expectation.. 3. The Population Variance is a Statistical Expectation.. 4. The Bernoulli Distribution.. 5. Introduction to Factorials and Combinatorials 6. The Binomial Distribution.. 7. Illustration of the Binomial Distribution.. 8. Resources for the Binomial Distribution
2 PubHlth Bernoulli and Binomial Page 2 of Review What is a Discrete Probability Distribution For a more detailed review, see Unit 2 Introduction to Probability, pp 56. Previously, we saw that A discrete probability distribution is a roster comprised of all the possibilities, together with the likelihood of the occurrence of each. The roster of the possibilities must comprise ALL the possibilities (be exhaustive) Each possibility has a likelihood of occurrence that is a number somewhere between zero and one. Looking ahead We ll have to refine these notions when we come to speaking about continuous distributions as, there, the roster of all possibilities is an infinite roster. Recall the Example of a Discrete Probability Distribution on pp 56 of Unit 2. We adopted the notation of using capital X as our placeholder for the random variable X = gender of a student selected at random from the collection of all possible students at a given university We adopted the notation of using little x as our placeholder for whatever value the random variable X might have x = 0 when the gender is male 1 when the gender is female x, generically.
3 PubHlth Bernoulli and Binomial Page 3 of 19 Value of the Random Variable X is x 0 = male 1 = female Probability that X has value x is Probability [ X = x ] Note that this roster exhausts all possibilities. Note that the sum of these individual probabilities, because the sum is taken over all possibilities, is 100% or Previously introduced was some terminology, too. 1. For discrete random variables, a probability model is the set of assumptions used to assign probabilities to each outcome in the sample space. The sample space is the universe, or collection, of all possible outcomes. 2. A probability distribution defines the relationship between the outcomes and their likelihood of occurrence. 3. To define a probability distribution, we make an assumption (the probability model) and use this to assign likelihoods.
4 PubHlth Bernoulli and Binomial Page 4 of Statistical Expectation Statistical expectation was introduced for the first time in Appendix 2 of Unit 2 Introduction to Probability, pp A variety of wordings might provide a clearer feel for statistical expectation. Statistical expectation is the long range average. The statistical expectation of what the state of Massachusetts will pay out is the long range average of the payouts taken over all possible individual payouts. Statistical expectation represents an on balance, even if on balance is not actually possible. IF $1 has a probability of occurrence = 0.50 $5 has a probability of occurrence = 0.25 $10 has a probability of occurrence = 0.15 and $25 has a probability of occurrence = 0.10 THEN on balance, the expected winning is $5.75 because $5.75 = [$1](0.50) + [$5](0.25) +[$10](0.15) + [$25](0.10) Notice that the on balance dollar amount of $5.75 is not an actual possible winning
5 PubHlth Bernoulli and Binomial Page 5 of 19 What can the State of Massachusetts expect to pay out on average? The answer is a value of statistical expectation equal to $5.75. [ Statistical expectation = $5.75 ] = [$1 winning] (percent of the time this winning occurs=0.50) + [$5 winning] (percent of the time this winning occurs =0.25) + [$10 winning] (percent of the time this winning occurs = 0.15) + [$25 winning] ](percent of the time this winning occurs = 0.10) You can replace the word statistical expectation with net result, long range average, or, on balance. Statistical expectation is a formalization of this intuition. For a discrete random variable X (e.g. winning in lottery) Having probability distribution as follows: Value of X, x = P[X = x] = $ $ $ $ The realization of the random variable X has statistical expectation E[X]=μ μ = [x]p(x=x) all possible X=x In the likely winnings example, μ = $5.75
6 PubHlth Bernoulli and Binomial Page 6 of 19 We can just as easily talk about the long range value of other things, too. The idea of statistical expectation is NOT a restricted one. Example If a lottery ticket costs $15 to purchase, what can he/she expect to attain? Your intuition tells you that the answer to this question is $ $15 = $9.25, representing a $9.25 loss. The long range loss of $9.25 is also a statistical expectation. Here s how it works. We ll define Y = (winning ticket price) Thus, Y = profit Value of Y, y = P[Y=y] = $ 1  $15 = $ $ 5  $15 = $ $10  $15 = $ $25  $15 = +$ The realization of the loss random variable Y has statistical expectation E[Y]=μ Y μ = [y]p(y=y) = $9.25 Y all possible Y=y
7 PubHlth Bernoulli and Binomial Page 7 of The Population Variance is a Statistical Expectation To keep things simple, let s revisit the example of the random variable defined as the winnings in one play of the Massachusetts State Lottery. The random variable X is the winnings. Recall that this variable has possible values x=$1, $5, $10, and $25. The statistical expectation of X is μ = $5.75. Recall that this figure is what the state of Massachusetts can expect to pay out, on average, in the long run. What about the variability in X? In learning about population variance σ 2 for the first time, we understood this to be a measure of the variability of individual values in a population. The population variance σ 2 of a random variable X is the statistical expectation of the quantity [ X μ ] 2 For a discrete random variable X (e.g. winning in lottery) Having probability distribution as follows: Value of [Xμ] 2 = P[X = x] = [15.75] 2 = [5 5.75] 2 = [ ] 2 = [ ] 2 = The variance of X is the statistical expectation of [Xμ] 2 = E ( X ) = [( x ) ]P(X=x) σ μ μ all possible X=x In the likely winnings example, σ 2 = dollars squared.
8 PubHlth Bernoulli and Binomial Page 8 of The Bernoulli Distribution Note The next 3 pages are nearly identical to pages of Unit 2, Introduction to Probability. They are reproduced here for ease of reading.  cb. The Bernoulli Distribution is an example of a discrete probability distribution. It is an appropriate tool in the analysis of proportions and rates. Recall the coin toss chance of heads can be recast as a random variable. Let Z = random variable representing outcome of one toss, with Z = 1 if heads 0 if tails π= Probability [ coin lands heads }. Thus, π= Pr [ Z = 1 ] We have what we need to define a probability distribution. Enumeration of all possible outcomes  outcomes are mutually exclusive  outcomes are exhaust all possibilities 1 0 Associated probabilities of each  each probability is between 0 and 1  sum of probabilities totals 1 Outcome Pr[outcome] 0 (1  π) 1 π
9 PubHlth Bernoulli and Binomial Page 9 of 19 In epidemiology, the Bernoulli might be a model for the description of ONE individual (N=1): This person is in one of two states. He or she is either in a state of: 1) event with probability π Recall the event might be mortality, MI, etc 2) non event with probability (1π) The model (quite a good one, actually) of the likelihood of being either in the event state or the nonevent state is given by the Bernoulli distribution Bernoulli Distribution Suppose Z can take on only two values, 1 or 0, and suppose: Probability [ Z = 1 ] = π Probability [ Z = 0 ] = (1π) This gives us the following expression for the likelihood of Z=z. Probability [ Z = z ] = π z (1π) 1z for z=0 or 1. Expected value (we call this μ) of Z is E[Z] = π Variance of Z (we call this σ 2 ) is Var[Z] = π (1π) Example: Z is the result of tossing a coin once. If it lands heads with probability =.5, then π =.5. Later, we ll see that individual Bernoulli distributions are the basis of describing patterns of disease occurrence in a logistic regression analysis.
10 PubHlth Bernoulli and Binomial Page 10 of 19 Mean (μ) and Variance (σ 2 ) of a Bernoulli Distribution Mean of Z = μ = π The mean of Z is represented as E[Z]. E[Z] = π because the following is true: E[Z] = [z]probability[z = z] All possible z = [0]Pr[Z=0]+[1]Pr[Z=1] = [ 0]( 1 π ) + [ 1]( π ) = π Variance of Z = σ 2 = (π)(1π) The variance of Z is Var[Z] = E[ (Z (EZ) 2 ]. Var[Z] = π(1π) because the following is true: 2 2 Var[Z] = E[(Z  π ) ] = [(z  π ) ]Probability[Z = z] All possible z 2 2 = [(0  π) ]Pr[Z = 0] +[(1 π ) ]Pr[Z = 1] = [ π 2 ]( 1 π) + [( 1 π) 2 ]( π) = π ( 1 π )[ π + ( 1 π )] = π ( 1 π )
11 PubHlth Bernoulli and Binomial Page 11 of Introduction to Factorials and Combinatorials A factorial is just a secretarial shorthand. Example  3! = (3)(2)(1) = 6 Example  8! = (8)(7)(6)(5)(4)(3)(2)(1) = 40,320 n! = (n)(n1)(n2) (3)(2)(1) Notice that the left hand side requires much less typesetting. We agree that 0! = 1 n factorial n! = (n)(n1)(n2) (2)(1) A combinatorial speaks to the question how many ways can we choose, without replacement and without regard to order? How many ways can we choose x=2 letters without replacement from the N=3 contained in { A, B, C }? By brute force, we see that there are 3 possible ways: AB AC BC Notice that the choice AB is counted once as it represents the same result whether the order is AB or BA More formally, how many ways can we choose without replacement is solved in two steps.
12 PubHlth Bernoulli and Binomial Page 12 of 19 Step 1 Ask the question, how many ordered selections of x=2 are there from N=3? By brute force, we see that there are 6 possible ordered selections: AB BA AC CA BC CB We can get 6 by another approach, too. 6 = (3 choices for first selection) x ( 31=2 choices for second selection) x (1 choice left) (3)(2) = (3)(2)(1) = 3! = 6 Following is a general formula for this idea # ordered selections of size x from N, without replacement, is (N)(N1)(N2) (Nx+1) = N! ( Nx )! Example  Applying this formula to our example, use N=3 and x=2, N! 3! 3! (3)(2)(1) = = = = 6 Nx! 3 2! 1! (1) ( ) ( ) Step 2 Correct for the multiple rearrangements of like results Consider the total count of 6. It counts the net result {AB} twice, once for the order AB and once for the order BA. This is because there are 2 ways to put the net result {AB} into an order (two choices for position 1 followed by one choice for position 2). Similarly, for the net result {AC} and for the net result {BC}. Thus, we want to divide by 2 to get the correct answer AB BA AC CA BC CB
13 PubHlth Bernoulli and Binomial Page 13 of 19  Here s another example. The number of ways to order the net result (WXYZ) is (4)(3)(2)(1) =24 because there are (4) choices for position #1 followed by (3) choices for position #2 followed by (2) choices for position #2 followed by (1) choice for position #1. # rearrangements (permutations) of a collection of x things is (x)(x1)(x2) (2)(1) = x! Now we can put the two together to define what is called a combinatorial. The combinatorial N x is the shorthand for the count of the # selections of size x, obtained without replacement, from a total of N N = x # ordered selections of size x correction for multiple rearrangements of x things = = N! ( Nx )! x! N! (Nx)!x!
14 PubHlth Bernoulli and Binomial Page 14 of The Binomial Distribution The Binomial is an extension of the Bernoulli. A Bernoulli can be thought of as a single event/nonevent trial. Now suppose we up the number of trials from 1 to N. The outcome of a Binomial can be thought of as the net number of successes in a set of N independent Bernoulli trials each of which has the same probability of event π. We d like to know the probability of X=x successes in N separate Bernoulli trials, but we do not care about the order of the successes and failures among the N separate trials. E.g.  What is the probability that 2 of 6 graduate students are female?  What is the probability that of 100 infected persons, 4 will die within a year? Steps in Calculating a Binomial Probability N = # of independent Bernoulli trials We ll call these trials Z 1, Z 2, Z N π = common probability of event accompanying each of the N trials. Thus, Prob [Z 1 = 1] = π, Prob [Z 2 = 1] = π, Prob [Z N = 1] = π π x (1π) Nx = Probability of one representative sequence that yields a net of x events and Nx nonevents. F HG E.g. A bent coin lands heads with probability =.55 and tails with probability =.45 Probability of sequence {HHTHH} = (.55)(.55)(.45)(.55)(.55) = [.55] 4 [.45] 1 N x I KJ = # ways to choose x from N
15 PubHlth Bernoulli and Binomial Page 15 of 19 Thus, Probability [N trials yields x events] = (# choices of x items from N) (Pr[one sequence]) N x π x ( π ) F = H G I K J 1 N x Formula for a Binomial Probability If a random variable X is distributed Binomial (N, π ) where N = # trials π = probability of event occurrence in each trial (common) Then the probability that the N trials yields x events is given by Pr [X = x] = N x x Nx π (1π)
16 PubHlth Bernoulli and Binomial Page 16 of 19 A Binomial Distribution is the sum of Independent Bernoulli Random Variables The Binomial is a summary of N individual Bernoulli trials Z i. Each can take on only two values, 1 or 0: Now consider N trials: Pr [ Z i = 1 ] = π for every individual Pr [ Z i = 0 ] = (1π) for every individual Among the N trials, what are the chances of x events? ( Z i = x )? The answer is the product of 2 terms. 1 st term: # selections of size x from a collection of N 2 nd term: Pr [ (Z 1 =1) (Z x =1) (Z x+1 =0) (Z N =0) ] This gives us the following expression for the likelihood of Z i = x: F HG I N Probability [ Z i = x ] = i= 1 Expected value is E[ Variance is Var[ N i= 1 N i= 1 N = # ways to choose X from N = x KJ F HG N x N i= 1 N i= 1 I π KJ x (1π) Nx for x=0,, N Z i = x] = N π Z i = x] = N π (1π) N! x!( N x)! where N! = N(N1)(N2)(N3) (4)(3)(2)(1) and is called the factorial. The Binomial is a description of a SAMPLE (Size = N): Some experience the event. The rest do not.
17 PubHlth Bernoulli and Binomial Page 17 of Illustration of the Binomial Distribution A roulette wheel lands on each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 with probability =.10. Write down the expression for the calculation of the following. #1. The probability of 5 or 6 exactly 3 times in 20 spins. #2. The probability of digit greater than 6 at most 3 times in 20 spins.
18 PubHlth Bernoulli and Binomial Page 18 of 19 Solution for #1. The event is an outcome of either 5 or 6 Thus, Probability [event] = π = spins says that the number of trials is N = 20 Thus, X is distributed Binomial(N=20, π=.20) F Pr[X = 3] = 20 3 HG I KJ = F H G I K J = [. 20] [. 80] 7 Solution for #2. The event is an outcome of either 7 or 8 or 9 Thus, Pr[event] = π =.30 As before, N = 20 Thus, X is distributed Binomial(N=20, π=.30) Translation: At most 3 times is the same as saying 3 times or 2 times or 1 time or 0 times which is the same as saying less than or equal to 3 times Pr[X 3] = Pr[X = 0] + Pr[X = 1] + Pr[X = 2] + Pr[X = 3] RF = S H G T IU KJ V W 20 3 x. 30 [. 70] 20 x= 0 x x 20 F = H G I K J F + H G I K J F + H G I K J F + H G I K J [. ] [. ] [. ] [. ] =.10709
19 PubHlth Bernoulli and Binomial Page 19 of Resources for the Binomial Distribution Note  To link directly to these resources, visit the BE course web site (wwwunix.oit.umass.edu/~biep540w). From the welcome page, click on BERNOULLI AND BINOMIAL DISTRIBUTIONS at left. Additional Reading A 2 page lecture on the Binomial Distribution from University of North Carolina. A very nice resource on the Binomial Distribution produced by Penn State University. Calculation of Binomial Probabilities Vassar Stats Exact Probability Calculator
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